longest-substring-with-at-least-k-repeating-characters

https://leetcode.com/problems/longest-substring-with-at-least-k-repeating-characters/

public class Solution {

public int longestSubstring(String s, int k) {

// Find count of each char

HashMap mp = new HashMap();

Object intObj;

for (int i=0; i<s.length(); i++) {

char ch = s.charAt(i);

intObj = mp.remove(ch);

int st = 0;

if (intObj != null) {

st = (int) intObj;

}

st++;

mp.put(ch, st);

}

// prepre iterate secondly

int ret = 0;

int last = -1;

HashMap newMp = new HashMap();

// iterate secondly

for (int i=0; i<s.length(); i++) {

char ch = s.charAt(i);

int num = (int)mp.get(ch);

// pass if num fits

if (num >= k) {

intObj = newMp.get(ch);

int newNum = 0;

if (intObj != null) {

newNum = (int) intObj;

}

newNum++;

newMp.put(ch, newNum);

continue;

}

// handle if meets nofit char

Set filter = new HashSet();

Iterator iter = newMp.entrySet().iterator();

Map.Entry entry;

// check newMp and prepare filter

while (iter.hasNext()) {

entry = (Map.Entry) iter.next();

char cch = (char)entry.getKey();

int cnt = (int)entry.getValue();

if (cnt < k) {

filter.add(cch);

}

int allCnt = (int)mp.remove(cch);

allCnt -= cnt;

mp.put(cch, allCnt);

}

// Prune

if (filter.size() == newMp.size()) {

last = i;

newMp.clear();

continue;

}

// use filter to check each segment

HashMap fMp = new HashMap();

int newLast = last;

for (int j=last+1; j<=i; j++) {

char fch = ' ';

if (j < i) {

fch = s.charAt(j);

}

// need to check segment

if (j == i || filter.contains(fch)) {

iter = fMp.entrySet().iterator();

// check map of each segment

while (iter.hasNext()) {

entry = (Map.Entry) iter.next();

char ffch = (char)entry.getKey();

int fcnt = (int)entry.getValue();

// Prepare Prune by update newMp

int newCnt = (int)newMp.remove(ffch);

newCnt -= fcnt;

newMp.put(ffch, newCnt);

if (newCnt < k) {

filter.add(ffch);

}

newLast = j;

fMp.clear();

// Check Prune

if (filter.size() == newMp.size()) {

break;

}

// no need to check segment, pass

else {

intObj = fMp.get(fch);

int fNum = 0;

if (intObj != null) {

fNum = (int) intObj;

}

fNum++;

fMp.put(fch, fNum);

if (fNum >= k) {

iter = fMp.entrySet().iterator();

boolean isFit = true;

while (iter.hasNext()) {

entry = (Map.Entry) iter.next();

int cfcnt = (int)entry.getValue();

if (cfcnt < k) {

isFit = false;

break;

}

if (isFit) {

if (j-newLast > ret) {

ret = j-newLast;

}

newMp.clear();

last = i;

}

if (s.length()-last-1 > ret) {

ret = s.length()-last-1;

}

return ret;

}

下面那个有未考虑到的地方，比如：

aaabbcbdcc

下面得出0，其实应该是3.

public class Solution {
    public int longestSubstring(String s, int k) {
        // Find count of each char
        HashMap mp = new HashMap();
        Object intObj;
        for (int i=0; i<s.length(); i++) {
            char ch = s.charAt(i);
            intObj = mp.remove(ch);
            int st = 0;
            if (intObj != null) {
                st = (int) intObj;
            }
            st++;
            mp.put(ch, st);
        }
        
        // prepre iterate secondly
        int ret = 0;
        int last = -1;
        HashMap newMp = new HashMap();
        
        // iterate secondly
        for (int i=0; i<s.length(); i++) {
            char ch = s.charAt(i);
            int num = (int)mp.get(ch);
            
            // pass if num fits
            if (num >= k) {
                intObj = newMp.get(ch);
                int newNum = 0;
                if (intObj != null) {
                    newNum = (int) intObj;
                }
                newNum++;
                newMp.put(ch, newNum);
                continue;
            }
            
            // handle if meets nofit char
            Set filter = new HashSet();
            Iterator iter = newMp.entrySet().iterator();
            Map.Entry entry;
            
            // check newMp and prepare filter
            while (iter.hasNext()) {
                entry = (Map.Entry) iter.next();
                char cch = (char)entry.getKey();
                int cnt = (int)entry.getValue();
                
                if (cnt < k) {
                    filter.add(cch);
                }
                
                int allCnt = (int)mp.remove(cch);
                allCnt -= cnt;
                mp.put(cch, allCnt);
            }
            
            // Prune
            if (filter.size() == newMp.size()) {
                last = i;
                newMp.clear();
                continue;
            }
            
            // use filter to check each segment
            HashMap fMp = new HashMap();
            int newLast = last;
            for (int j=last+1; j<=i; j++) {
                char fch = ' ';
                if (j < i) {
                    fch = s.charAt(j);
                }
                
                // need to check segment
                if (j == i || filter.contains(fch)) {
                    iter = fMp.entrySet().iterator();
                    boolean fits = true;
                    
                    // check map of each segment
                    while (iter.hasNext()) {
                        entry = (Map.Entry) iter.next();
                        char ffch = (char)entry.getKey();
                        int fcnt = (int)entry.getValue();
                                        
                        if (fcnt < k) {
                            fits = false;
                        }
                        
                        // Prepare Prune by update newMp                      
                        int newCnt = (int)newMp.remove(ffch);
                        newCnt -= fcnt;
                        newMp.put(ffch, newCnt);
                        
                        if (newCnt < k) {
                            filter.add(ffch);
                        }
                    }
                    
                    // update final ret
                    if (fits) {
                        if (j-newLast-1 > ret) {
                            ret = j-newLast-1;
                        }
                    }
                    newLast = j;
                    fMp.clear();
                    
                    // Check Prune
                    if (filter.size() == newMp.size()) {
                        break;
                    }
            
                }
                // no need to check segment, pass
                else {
                    intObj = fMp.get(fch);
                    int fNum = 0;
                    if (intObj != null) {
                        fNum = (int) intObj;
                    }
                    fNum++;
                    fMp.put(fch, fNum);
                }
            }
            newMp.clear();
            last = i;
            
        }
        if (s.length()-last-1 > ret) {
            ret = s.length()-last-1;
        }
        return ret;
    }
}


test case:

"zzzzzzzzzzaaaaaaaaabbbbbbbbhbhbhbhbhbhbhicbcbcibcbccccccccccbbbbbbbbaaaaaaaaafffaahhhhhiaahiiiiiiiiifeeeeeeeeee"
10

expected return: 21

来源：https://www.cnblogs.com/charlesblc/p/5879401.html

标签

char

last

hashmap